Transcript Thermal Physics

Thermal Physics
13 - Temperature & Kinetic Energy
15 - Laws of Thermodynamics
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Assignments
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Read 13.2,4,6-10
Read 14.1-4 for review
Read 15.1-9
Asgn
13/Q3,5,27 and P32,34,44
15/Q1-3 and P58,10,16,21,23,24,32,39,42
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• Introduction to Thermodynamics
•
•
Temperature and Heat
Gas Law Review
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Thermodynamics
• This is the study of heat and htermal
energy.
• Thermal properties (heat and temperature)
are based on the motion of molecules.
This is one part of our studies that relates
to Chemistry.
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Total Energy
• E = U + K + Eint
•
U: potential energy
•
K: kinetic energy
•
Eint: internal or thermal energy
• Potential and kinetic energies are
specifically for large objects and represent
mechanical energy
• Thermal energy relates to the kinetic
energy of the molecules of a substance.
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Temperature and Heat
• Temperature is a measure of the average kinetic
energy of the molecules of a substance - a
measure of how fast the molecules are moving.
Unit: oC or K
• Temperature is NOT heat!
• Heat is the internal energy that is transferred
between bodies in contact. Unit: joules (J) or
calories (cal)
• A difference in temperatures will cause heat
energy to be exchanged between bodies in
contact When two bodies are at the same
temperature, no heat is transferred; this is called
Thermal Equilibrium.
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Ideal Gas Law (combined Gas Law)
• P1V1/T1 = P2V2/T2
•
P1,P2: initial and final pressure (any unit)
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•
V1,V2: initial and final volume (any unit)
T1,T2: initial and final temperature (in Kelvin)
• Temperature in K = oC + 273.
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Sample Problem
• An ideal gas occupies 4.0 L at 23oC and
2.3 atm. What will be the volume of the
gas if the temperature is lowered to 0oC
and the pressure is 3.1 atm?
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Ideal Gas Equation
• PV = nRT (using moles)
• PV = N kB T (using molecules)
• P: pressure (Pa)
• V: volume (m3)
• n: number of moles
• N: number of molecules
• R: Universal Gas constant (8.31 J/mol K)
• kB: Boltzman’s constant (1.38 x 10-23 J/K)
• T: temperature (in K)
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Sample Problem
• Determine the number of moles of an ideal
gas that occupies 10.0 m3 at atmospheric
pressure and 25oC.
• PV = nRT
• n = PV/RT
• n = 100,00 Pa(10.0 m3)
•
8.31 J / molK (25+273K)
• n=
mol
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Sample Problem
• Suppose a near vacuum contains 25,000
molecules of helium in one cubic meter at
0oC. What is the pressure?
• PV = nRT
n =N
molecules
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NA Avogadro’s #
• PV = NkBT
• PV = NkBT = 25,000(1.38x10-23J/K)(273K)
•
V
• P = 9.42 x 10-17 Pa
1m3
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The relationship
• between the Universal Gas Law Constant,
R (8.31 J/mol K) and Boltzman’s constant
kB (1.38 x 10 -23 J/K) is…
• Na kB = R
• 6.02 x 1023 (1.38 x 10-23 J/K) = 8.32 J/molK
•
mole
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Kinetic Theory of Gases
• 1. Gases consist of a large number of
molecules that make elastic collisions with
each other and the walls of the container.
• 2. Molecules are separated, on average,
by large distances and exert no forces on
each other except when they collide.
• 3. There is no preferred position for a
molecule in the container, and no
preferred direction for the velocity.
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Average Kinetic Energy of a Gas
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Kave = 3/2 kB T
Kave : average kinetic energy (J)
kB : Boltzmann’s constant (1.38 x 10-23 J/K)
T: Temperature (K)
The molecules have a range of kinetic
energies; is just the average of that range.
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Sample Problem
• What is the average kinetic energy and the
average speed of oxygen molecules in a
gas sample at OoC?
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Sample Problem
• Suppose nitrogen and oxygen are in a sample of
gas at 100oC.
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A. What is the ratio of the average kinetic
energies for the two molecules?
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B. What is the ratio of their average speeds?
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Notation Warning!
• U is potential energy in mechanics.
However,
• U is Eint (thermal energy) in
thermodynamics
• This means that when we are in thermo, U
is thermal energy which is related to
temperature. When in mechanics, it is
potential energy (related to configuration
or position).
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More about U
• U is the sum of the
• Heat added is +
kinetic energies of all
molecules in a system • Heat lost is (or gas).
• U = N Kave
• Work on system is • U = N (3/2 kB T)
• U = n (3/2 R T)
• Work by system is +
– Since kB = R/NA
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First Law of Thermodynamics
 DU = Q - W
 DU : change in internal energy of system (J)
 Q: heat added to the system (J). This heat
exchange is driven by temperature difference.
 W: work done on the system (J). Work will be
related to the change in the system’s volume.
 This law is sometimes paraphrased as “you
can’t win”
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Problem #8
DU = Q - W
DU = 200 J - 100 J
Heat added is +
Heat lost is -
DU = 100 J
Work on system is Work by system is +
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Problem #9
DU = Q - W
45,000 J - 40,000 J = 0 - W
W = - 5,000 J
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Gas Processes
• The thermodynamic state of a gas is
defined by pressure, volume, and
temperature.
• A “gas process” describes how gas gets
from one state to another state.
• Processes depend on the behavior of the
boundary and the environment more than
they depend on the behavior of the gas.
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D U = 0. Q = W
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Isobaric
Isometric/isochoric
P is constant
V is constant
W=PDV
W=0
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Work
• Calculation of work done on a system (or
by a system) is an important part of
thermodynamic calculations.
• Work depends upon volume change
• Work also depends upon the pressure at
which the volume change occurs.
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Problem #10
W = P D V = 1 atm (.8-.02) m3
W = 1.013x105 N/m2 (.78m3)
W=+
work done by a gas
W=-
work done on a gas
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Problem #11
U=Q-W
230 = -120 - W
W = -350J
W = P D V  D V =W/P =
-350 J / 1.013 x 105 J/m3
DV=
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Work - isobaric
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Work is path dependent
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Problem #12
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Work done by a Gas
When a gas undergoes a complete cycle, it
starts and ends in the same state. The gas
is identical before and after the cycle.
There is no identifiable change in the gas.
D U = 0 for a complete cycle.
However, the environment, has been
changed.
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Work done by the Gas
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#13 Problem
• Consider the cycle ABCDA, where
– State A: 200 kPa, 1.0 m3
– State B: 200 kPa, 1.5 m3
– State C: 100 kPa, 1.5 m3
– State D: 100 kPa, 1.0 m3
 A. Sketch the cycle.
 B. Graphically estimate the work done by the
gas in one cycle.
 C. Estimate the work done by the
environment in one cycle.
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A: 200 kPa, 1.0 m3
C: 100 kPa, 1.5 m3
B: 200 kPa, 1.5 m3
D: 100 kPa, 1.0 m3
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• 10.(II) Consider the following two-step
process. Heat is allowed to flow out of an
ideal gas at constant volume so that its
pressure drops from 2.2 atm to 1.4 atm.
Then the gas expands at constant
pressure, from a volume of 6.8 L to 9.3 L,
where the temperature reaches its original
value. See Fig. 15–22. Calculate (a) the
total work done by the gas in the process,
(b) the change in internal energy of the
gas in the process, and (c) the total heat
flow into or out of the gas.
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Problem 15/10
Fig. 15-22
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• (a) No work is done during the first step, since the
volume is constant. The work in the second step
• is given by
 1.01 105 Pa 
1 10-3 m3
2
W  PDV  1.4 atm  
9.3L
6.8
L

3.5

10
J



1L
 1 atm 
(b) Since there is no overall change in temperature,
DU  0 J
(c) The heat flow can be found from the first law of
thermodynamics.
DU  Q - W  Q  DU  W  0  3.5  102 J= 3.5  10 2 J  into the gas 
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2nd Law of Thermodynamics
• No process is possible whose sole result
is the complete conversion of heat from a
hot reservoir into mechanical work.
(Kelvin-Planck statement.)
• No process is possible whose sole result
is the transfer of heat from a colder to a
hotter body. (Clausius statement.)
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Heat Engines
• Heat Engines
• The Carnot Cycle
• Efficiency
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Heat Engines
• Heat engines can convert heat into useful
work.
• According to the 2nd Law of T-D, heat engines
always produce some waste heat.
• Efficiency can be used to tell how much heat
is needed to produce a given amount of
work.
• NOTE: A heat engine is not something that
produces heat. A heat engine transfers heat
from hot to cold, and does mechanical work
in the process.
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QH = W + QL
Operating Temps:
QH and QL
New Sign
Convention:
QH, QL and W are all
positive
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Efficiency of Heat Engine
• In general, efficiency is related by what
fraction of the energy put into a system is
convereted to useful work.
• In the case of a heat engine, the energy
that is put in is the heat that flows into the
system from the hot reservoir.
• Only some of the heat flowing in is
converted to work. The rest if waste heat
that is dumped into the cold reservoir.
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Efficiency of Heat Engine
• Efficiency = W/QH = (QH - QC)/QH *
• W: Work done by engine on enviroment
• QH: Heat absorbed from hot reservoir
• QC: Waste heat dumped to cold reservoir
• Efficiency is often given as percent
efficiency.
•
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Adiabatic vs Isothermal Expansion
In adiabatic
expansion, no heat
energy can enter
the gas to replace
energy being lost as
it does work on the
environment. The
temperature drops,
and so does the
pressure.
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Carnot Cycle
Go to page 419 in your book.
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Efficiency of Carnot Cycle
• For a Carnot engine, the efficiency can be
calculated from the temperatures of the
hot and cold reservoirs.
• Carnot Efficiency: e = 1 - TC
•
TH
• TC:Temperature of cold reservoir (K)
• TH:Temperature of hot reservoir (K)
•
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Sample Problem
• 23. (II) A Carnot engine performs work at the rate of 440
kW while using 680 kcal of heat per second. If the
temperature of the heat source is 570°C, at what
temperature is the waste heat exhausted?
TL
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Sample Problem
• 21. (II) A nuclear power plant operates at 75% of its
maximum theoretical (Carnot) efficiency between
temperatures of 625°C and 350°C. If the plant produces
electric energy at the rate of 1.3 GW, how much exhaust
heat is discharged per hour?
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Entropy
• Disorder or randomness
• The entropy of the universe is increasing.
This will lead to what philosophers have
often discussed as “Heat Death of the
Universe.
• Does the entropy of your room tend to
increase or decrease?
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Entropy
 DS = Q/T
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DS: Change in entropy (J/K)
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Q: Heat going into system (J)
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T: Kelvin temperature (K)
• If change in entropy is positive, randomness or
disorder has increased.
• Spontaneous changes involve an increase in
entropy.
• Generally, entropy can go down only when
energy is put into the system.
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Problem 15/42
• 42. (II) 1.0 kg of water at 30°C is mixed with 1.0
kg of water at 60°C in a well-insulated container.
Estimate the net change in entropy of the
system.
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